Calculus 10th Edition

by
Larson, Ron; Edwards, Bruce H.

Published by
Brooks Cole

ISBN 10:
1-28505-709-0

ISBN 13:
978-1-28505-709-5

Chapter 2 - Differentiation - 2.2 Exercises - Page 115: 64

Answer

$k = -\frac{1}{3}$

Work Step by Step

To reverse engineer this type of exercise, one must remember what is needed to generate the equation of a line tangent to a function: (1) the point $(x,y)$ of interest and (2) the slope of the line, as calculated from the function's first derivative.
We know that, at the point where the line is tangent to the function, both functions share $(x,y)$. As such, we can equate both of them: $$kx^{2} = -2x + 3$$ which we can rewrite as follows: $$kx^{2} + 2x - 3 = 0$$ Using the quadratic formula, $$x = \frac{-b +/- (b^{2} - 4ac)^{\frac{1}{2}}}{2a}$$ we can solve for x: $$x = \frac{-2 +/- (2^{2} - 4(k)(-3))^{\frac{1}{2}}}{2(k)}$$ $$x = \frac{-1 +/- (1 + 3k)^{\frac{1}{2}}}{k}$$
Furthermore, we also know that the first derivative of the function $f(x)$ is the equation for the slope of the tangent line: $$f(x) = kx^{2}$$ $$f'(x) = 2kx$$ We know that the slope must be equal to -2 because of the equation for the tangent line that was given: $$-2 = 2kx$$ $$-1 = kx$$ And now we can plug in the previous value for x that we calculated: $$-1 = k * \frac{-1 +/- (1 + 3k)^{\frac{1}{2}}}{k}$$ $$-1 = -1 +/- (1 + 3k)^{\frac{1}{2}}$$ $$0 = [+/- (1 + 3k)^{\frac{1}{2}}]^{2}$$ $$0 = 1 + 3k$$ Therefore, $k = -\frac{1}{3}$